Completed Binary Search-1

Problem-6 RotatedSortedArray.py

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+# 33. Search in Rotated Sorted Array
+
+# Time Complexity: O(logn)
+# Space Complexity: O(1)
+
+"""
+Logic: Using Binary search reject one half of the input where the target cannot lie. 
+Note: Atleast one half of a rotated sorted array in always sorted. 
+"""
+
+class Solution(object):
+    def search(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        if not nums:
+            return -1
+
+        l = 0
+        h = len(nums) - 1
+
+        while l <= h:
+            mid = (l+h)//2
+
+            if nums[mid] == target:
+                return mid
+
+            if nums[l] <= nums[mid]:
+                if nums[l] <= target <= nums[mid]:
+                    h = mid - 1
+                else:
+                    l = mid + 1
+            else:
+                if nums[mid] <= target <= nums[h]:
+                    l = mid + 1
+                else:
+                    h = mid - 1
+
+        return -1

Problem-7 SearchArrayOfUnknownSize.py

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+# 702. Search in a Sorted Array of Unknown Size
+
+# """
+# This is ArrayReader's API interface.
+# You should not implement it, or speculate about its implementation
+# """
+#class ArrayReader(object):
+#    def get(self, index):
+#        """
+#        :type index: int
+#        :rtype int
+#        """
+
+class Solution(object):
+    def search(self, reader, target):
+        """
+        :type reader: ArrayReader
+        :type target: int
+        :rtype: int
+        """
+        l = 0
+        h = 1
+
+        # Find the range of index where the target maybe present
+        while target > reader.get(h):
+            l = h
+            h = h*2
+
+        # In the range using binary search, try to find the target
+        while l <= h:
+            mid = l + (h-l)//2
+
+            if target == reader.get(mid):
+                return mid
+            elif target > reader.get(mid):
+                l = mid + 1
+            else:
+                h = mid - 1
+
+        return -1

Problem-8 Search2DMatrix.py

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+# 74. Search a 2D Matrix
+
+# Time Complexity: O(log(mn))
+# Space Complexity: O(1)
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+
+        l = 0
+        h = len(matrix) * len(matrix[0]) - 1
+
+        while l <= h:
+            mid = l + (h-l) //2
+
+            r = mid/len(matrix[0])
+            c = mid%len(matrix[0])
+
+            if(matrix[r][c] == target):
+                return True
+            elif(matrix[r][c] > target):
+                h = mid - 1
+            else:
+                l = mid + 1
+        return False